Let's break down the problem step by step and address each question in sequence.

1. Normal force pushing the ladder up from the floor

The normal force $F_N$ from the floor must balance the weight of the ladder, which acts directly downward due to gravity. The weight of the ladder is given by: $W = mg$ where:

• $m = 35 \, \text{kg}$ is the mass of the ladder, and
• $g = 9.81 \, \text{m/s}^2$ is the acceleration due to gravity.

So the weight is: $W = 35 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 343.35 \, \text{N}$ The normal force from the floor is equal to the weight: $F_N = 343.35 \, \text{N}$

2. Torque exerted by the weight of the ladder

The torque $\tau$ exerted by the ladder's weight, using the floor as the pivot point, is calculated using the formula for torque: $\tau = r F \sin(\theta)$ where:

• $r = \frac{L}{2} = \frac{3.6 \, \text{m}}{2} = 1.8 \, \text{m}$ is the distance from the pivot to the center of mass of the ladder,
• $F = W = 343.35 \, \text{N}$ is the weight of the ladder, and
• $\theta = 65^\circ$ is the angle between the ladder and the floor.

Thus, the torque is: $\tau = 1.8 \, \text{m} \times 343.35 \, \text{N} \times \sin(65^\circ)$ $\tau \approx 1.8 \times 343.35 \times 0.9063 = 560.51 \, \text{N} \cdot \text{m}$

3. Normal force from the wall

To balance the torque caused by the weight, the normal force from the wall $F_w$ must create an equal and opposite torque. The torque due to the normal force is given by: $\tau_w = F_w \cdot L \cdot \sin(90^\circ - 65^\circ) = F_w \cdot 3.6 \cdot \sin(25^\circ)$ Since the torques must balance, we set $\tau_w = \tau$: $F_w \cdot 3.6 \cdot \sin(25^\circ) = 560.51$ Solving for $F_w$: $F_w = \frac{560.51}{3.6 \times 0.4226} = \frac{560.51}{1.52136} \approx 368.27 \, \text{N}$

4. Minimum coefficient of friction to keep the ladder from slipping

The normal force from the wall $F_w$ must be balanced by the friction force $F_f$ from the floor: $F_f = F_w = 368.27 \, \text{N}$ The friction force is also related to the normal force and the coefficient of friction $\mu$: $F_f = \mu F_N$ Thus, the coefficient of friction is: $\mu = \frac{F_f}{F_N} = \frac{368.27}{343.35} \approx 1.07$

Summary of results:

1. Normal force from the floor: $F_N = 343.35 \, \text{N}$
2. Torque exerted by the ladder's weight: $\tau = 560.51 \, \text{N} \cdot \text{m}$
3. Normal force from the wall: $F_w = 368.27 \, \text{N}$
4. Minimum coefficient of friction: $\mu = 1.07$

Would you like further clarification or more details on any step? Here are 5 related questions for practice:

1. How would the results change if the angle of the ladder were increased to 75°?
2. What happens to the normal force from the wall if the mass of the ladder doubles?
3. Can you calculate the friction force if the coefficient of friction was 0.9 instead?
4. How would the torque change if the ladder’s length was 5 meters instead of 3.6 meters?
5. What role does the center of mass play in determining the torque exerted by the ladder?

Tip: The torque is highly dependent on the angle between the force and the lever arm. Adjusting the angle can significantly change the resulting torque.